60 Ibs. per.sq. ft. (according to the exposure) upon the area
(k x width of bay).
Total force Pt = 56 x k wf, say, in Ibs.
Then Pt may be resolved into two forces, one parallel to the
rafter, and one (Pn) normally, and
Pn total = ~—- ff
This force is distributed at a, d, and c as -fV .^n> f Pn, A Pn.
In iron roofs the expansion is usually allowed for by fixing one
end (a) and leaving the other free (£), which allows us to say the
reaction, Rt2, is vertical. Now Rtl, Rt25 Pn are three balancing
forces, and must meet in one point X, found by producing Pn to
meet Rt2 produced. Then joining a X, the direction of Rtl is
found, and the amounts Rtl -and Rt2 further obtained from the
auxiliary diagram. If the wind blow from the right, Pn is exerted
on cb, and x will be above instead of below b (Case III.). Both
II. and III. must be examined, though we only, have space for
Case II. Take the lettering as in L, with the exception of the
additional external space L, and draw the stress diagrams as
Finally, tabulate the stresses for Cases L, II., and III.: then
add L to II. and III. separately to find the maximum stresses,
and design.th'e members to suit. (See Appendix, IV,, p. 953.)
Framed Structures of Three Dimensions are such
as include a solid instead of an area. They must be solved by a
step-by-step process, taking each plane in succession. We will
explain by means of an example.
Example 39.—A sheer legs (Fig. 433) is formed of two fore legs
145' long and 60' apart at the base, and a back leg- 170' long attached
to a nut having a travel of 40'. The maximum overhang is 40', and
load too tons. Find the stresses in the members, (i) and (2), at each
end of the nut stroke; (3) when the load'is directly over the base
plates, (Hons. Mach, Gonstr. Ex.'1888.)
Case L Nut at P.—Taking first the plane AD B, the diagram P
shews stresses . A _,
Turning to the end view, the stress of 174 in AB must be resolved in
each leg, as in diagram Q, giving stresses
in a b and a c, each =* 88 tons.